Answer
F = 2.97 N
Work Step by Step
Let's consider the system of block A. Since the speed is constant, the tension $T$ is equal in magnitude to the force of kinetic friction exerted on block A.
$T = m_A~g~\mu_k$
Let's consider the horizontal forces acting on block B.
$F = (m_A+m_B)~g~\mu_k + m_A~g~\mu_k + T$
$F = (m_A+m_B)~g~\mu_k + m_A~g~\mu_k + m_A~g~\mu_k$
$F = m_B~g~\mu_k + 3~m_A~g~\mu_k$
$F = (4.20~N)(0.30) + (3)(1.90~N)(0.30)$
$F = 2.97~N$
